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Chapter 1
INTRODUCTION AND BASIC CONCEPTS
Thermodynamics
1-1C Classical thermodynamics is based on experimental observations whereas statistical thermodynamics is based on the average behavior of large groups of particles.
1-2C On a downhill road the potential energy of the bicyclist is being converted to kinetic energy, and thus the bicyclist picks up speed. There is no creation of energy, and thus no violation of the conservation of energy principle.
1-3C There is no truth to his claim. It violates the second law of thermodynamics.
1-4C A car going uphill without the engine running would increase the energy of the car, and thus it would be a violation of the first law of thermodynamics. Therefore, this cannot happen. Using a level meter (a device with an air bubble between two marks of a horizontal water tube) it can shown that the road that looks uphill to the eye is actually downhill.
Mass, Force, and Units
1-5C Pound-mass lbm is the mass unit in English system whereas pound-force lbf is the force unit. One pound-force is the force required to accelerate a mass of 32.174 lbm by 1 ft/s^2. In other words, the weight of a 1-lbm mass at sea level is 1 lbf.
1-6C In this unit, the word light refers to the speed of light. The light-year unit is then the product of a velocity and time. Hence, this product forms a distance dimension and unit.
1-7C There is no acceleration, thus the net force is zero in both cases.
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1-8E The weight of a man on earth is given. His weight on the moon is to be determined.
Analysis Applying Newton's second law to the weight force gives
4 lbm 1 lbf
174 lbmft/s
10 ft/s
180 lbf^2 2 ⎟=
g
W
W mg m
Mass is invariant and the man will have the same mass on the moon. Then, his weight on the moon will be
⎟= 30.7^ lbf ⎠
= =^22
- 174 lbmft/s
1 lbf W mg ( 180. 4 lbm)( 5. 47 ft/s )
1-9 The interior dimensions of a room are given. The mass and weight of the air in the room are to be determined.
Assumptions The density of air is constant throughout the room.
Properties The density of air is given to be ρ = 1.16 kg/m^3.
ROOM AIR
6X6X8 m^3
Analysis The mass of the air in the room is
m = ρ V=(1.16 kg/m^3 )(6× 6 × 8 m^3 )= 334.1 kg
Thus,
⎟⎟=^3277 N
= =^22
1 kg m/s
1 N
W mg (334.1kg)(9.81m/s)
1-10 The variation of gravitational acceleration above the sea level is given as a function of altitude. The height at which the weight of a body will decrease by 1% is to be determined.
Sea level
z
Analysis The weight of a body at the elevation z can be expressed as
W = mg = m ( .9 807 − 3 32. × 10 −^6 z )
In our case,
W = 0 99. W (^) s = 0 99. mg (^) s =0 99. ( m )( .9 807 )
Substituting,
- 99 ( 9. 81 )= ( 9. 81 − 3. 32 × 10 −^6 z )⎯⎯→ z = 29,539 m
PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
1-14 EES Problem 1-13 is reconsidered. The entire EES solution is to be printed out, including the numerical results with proper units.
Analysis The problem is solved using EES, and the solution is given below.
W=m*g "[N]" m=5 [kg] g=9.79 [m/s^2]
"The force balance on the rock yields the net force acting on the rock as" F_net = F_up - F_down"[N]" F_up=150 [N] F_down=W"[N]"
"The acceleration of the rock is determined from Newton's second law." F_net=a*m
"To Run the program, press F2 or click on the calculator icon from the Calculate menu"
SOLUTION a=20.21 [m/s^2] F_down=48.95 [N] F_net=101.1 [N] F_up=150 [N] g=9.79 [m/s^2] m=5 [kg] W=48.95 [N]
1-15 Gravitational acceleration g and thus the weight of bodies decreases with increasing elevation. The percent reduction in the weight of an airplane cruising at 13,000 m is to be determined.
Properties The gravitational acceleration g is given to be 9.807 m/s^2 at sea level and 9.767 m/s^2 at an altitude of 13,000 m.
Analysis Weight is proportional to the gravitational acceleration g , and thus the percent reduction in weight is equivalent to the percent reduction in the gravitational acceleration, which is determined from
× = 0.41%
× =
%Reduction in weight %Reductionin 100 g
g g
Therefore, the airplane and the people in it will weight 0.41% less at 13,000 m altitude.
Discussion Note that the weight loss at cruising altitudes is negligible.
PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
Systems, Properties, State, and Processes
1-16C This system is a region of space or open system in that mass such as air and food can cross its control boundary. The system can also interact with the surroundings by exchanging heat and work across its control boundary. By tracking these interactions, we can determine the energy conversion characteristics of this system.
1-17C The system is taken as the air contained in the piston-cylinder device. This system is a closed or fixed mass system since no mass enters or leaves it.
1-18C Carbon dioxide is generated by the combustion of fuel in the engine. Any system selected for this analysis must include the fuel and air while it is undergoing combustion. The volume that contains this air- fuel mixture within piston-cylinder device can be used for this purpose. One can also place the entire engine in a control boundary and trace the system-surroundings interactions to determine the rate at which the engine generates carbon dioxide.
1-19C When analyzing the control volume selected, we must account for all forms of water entering and leaving the control volume. This includes all streams entering or leaving the lake, any rain falling on the lake, any water evaporated to the air above the lake, any seepage to the underground earth, and any springs that may be feeding water to the lake.
1-20C Intensive properties do not depend on the size (extent) of the system but extensive properties do.
1-21C The original specific weight is
V
W
If we were to divide the system into two halves, each half weighs W /2 and occupies a volume of V/2. The
specific weight of one of these halves is
γ = = γ V
W
which is the same as the original specific weight. Hence, specific weight is an intensive property.
1-22C The number of moles of a substance in a system is directly proportional to the number of atomic particles contained in the system. If we divide a system into smaller portions, each portion will contain fewer atomic particles than the original system. The number of moles is therefore an extensive property.
1-23C For a system to be in thermodynamic equilibrium, the temperature has to be the same throughout but the pressure does not. However, there should be no unbalanced pressure forces present. The increasing pressure with depth in a fluid, for example, should be balanced by increasing weight.
PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
1-29 EES The variation of density of atmospheric air with elevation is given in tabular form. A relation for the variation of density with elevation is to be obtained, the density at 7 km elevation is to be calculated, and the mass of the atmosphere using the correlation is to be estimated.
Assumptions 1 Atmospheric air behaves as an ideal gas. 2 The earth is perfectly sphere with a radius of 6377 km, and the thickness of the atmosphere is 25 km.
Properties The density data are given in tabular form as
0 5 10 15 20 25
0
1
z, km
, kg/m
3
r , km z , km ρ, kg/m^3
Analysis Using EES, (1) Define a trivial function rho= a+z in equation window, (2) select new parametric table from Tables, and type the data in a two-column table, (3) select Plot and plot the data, and (4) select plot and click on “curve fit” to get curve fit window. Then specify 2nd^ order polynomial and enter/edit equation. The results are:
ρ( z ) = a + bz + cz^2 = 1.20252 – 0.101674 z + 0.0022375 z^2 for the unit of kg/m^3 ,
(or, ρ( z ) = (1.20252 – 0.101674 z + 0.0022375 z^2 )× 109 for the unit of kg/km^3 )
where z is the vertical distance from the earth surface at sea level. At z = 7 km, the equation would give ρ = 0.60 kg/m^3.
( b ) The mass of atmosphere can be evaluated by integration to be
4 [ ( 2 ) / 2 ( 2 ) / 3 ( 2 ) / 4 / 5 ]
4 5 0
2 3 0 0
2 0 0
2 0
2 0
2 0
2 0
2 0
2 0
ar h r a br h a br cr h b cr h ch
m dV a bz cz r z dz a bz cz r rz z dz
h z
h z V = + + + + + + + +
= =
where r 0 = 6377 km is the radius of the earth, h = 25 km is the thickness of the atmosphere, and a = 1.20252, b = -0.101674, and c = 0.0022375 are the constants in the density function. Substituting and multiplying by the factor 10^9 for the density unity kg/km^3 , the mass of the atmosphere is determined to be
m = 5.092 × 1018 kg
Discussion Performing the analysis with excel would yield exactly the same results.
EES Solution for final result:
a=1.2025166; b=-0. c=0.0022375; r=6377; h= m=4pi(ar^2h+r(2a+br)h^2/2+(a+2br+cr^2)h^3/3+(b+2cr)h^4/4+ch^5/5)*1E+
PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
Temperature
1-30C The zeroth law of thermodynamics states that two bodies are in thermal equilibrium if both have the same temperature reading, even if they are not in contact.
1-31C They are Celsius (°C) and kelvin (K) in the SI, and fahrenheit (°F) and rankine (R) in the English system.
1-32C Probably, but not necessarily. The operation of these two thermometers is based on the thermal expansion of a fluid. If the thermal expansion coefficients of both fluids vary linearly with temperature, then both fluids will expand at the same rate with temperature, and both thermometers will always give identical readings. Otherwise, the two readings may deviate.
1-33 A temperature is given in °C. It is to be expressed in K.
Analysis The Kelvin scale is related to Celsius scale by
T (K] = T (°C) + 273
Thus, T (K] = 37°C + 273 = 310 K
1-34E A temperature is given in °C. It is to be expressed in °F, K, and R.
Analysis Using the conversion relations between the various temperature scales,
T (K] = T (°C) + 273 = 18°C + 273 = 291 K T (°F] = 1.8 T (°C) + 32 = (1.8)(18) + 32 = 64.4 ° F
T (R] = T (°F) + 460 = 64.4 + 460 =^ 524.4 R
1-35 A temperature change is given in °C. It is to be expressed in K.
Analysis This problem deals with temperature changes, which are identical in Kelvin and Celsius scales. Thus, Δ T (K] = Δ T (°C) = 15 K
PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
Pressure, Manometer, and Barometer
1-40C The pressure relative to the atmospheric pressure is called the gage pressure , and the pressure relative to an absolute vacuum is called absolute pressure.
1-41C The blood vessels are more restricted when the arm is parallel to the body than when the arm is perpendicular to the body. For a constant volume of blood to be discharged by the heart, the blood pressure must increase to overcome the increased resistance to flow.
1-42C No, the absolute pressure in a liquid of constant density does not double when the depth is doubled. It is the gage pressure that doubles when the depth is doubled.
1-43C If the lengths of the sides of the tiny cube suspended in water by a string are very small, the magnitudes of the pressures on all sides of the cube will be the same.
1-44C Pascal’s principle states that the pressure applied to a confined fluid increases the pressure throughout by the same amount. This is a consequence of the pressure in a fluid remaining constant in the horizontal direction. An example of Pascal’s principle is the operation of the hydraulic car jack.
1-45E The maximum pressure of a tire is given in English units. It is to be converted to SI units.
Assumptions The listed pressure is gage pressure.
Analysis Noting that 1 atm = 101.3 kPa = 14.7 psi, the listed maximum pressure can be expressed in SI units as
⎟⎟=^241 kPa ⎠
14.7psi
- 3 kPa P max 35 psi ( 35 psi)
Discussion We could also solve this problem by using the conversion factor 1 psi = 6.895 kPa.
PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
1-46 The pressure in a tank is given. The tank's pressure in various units are to be determined.
Analysis Using appropriate conversion factors, we obtain
( a ) (^) ⎟⎟= 1500 kN/m^2 ⎠
1 kPa
1 kN/m ( 1500 kPa)
2 P
( b ) (^) ⎟⎟= 1,500,000 kg/m ⋅ s^2 ⎠
1 kN
1000 kg m/s 1 kPa
1 kN/m ( 1500 kPa)
2 2 P
( c ) (^) ⎟= 1,500,000, 000 kg/km ⋅ s^2 ⎠
1 km
1000 m 1 kN
1000 kgm/s 1 kPa
1 kN/m ( 1500 kPa)
2 2 P
1-47E The pressure given in kPa unit is to be converted to psia.
Analysis Using the kPa to psia units conversion factor,
⎟= 29.0^ psia ⎠
- 895 kPa
1 psia P ( 200 kPa)
1-48E A manometer measures a pressure difference as inches of water. This is to be expressed in psia unit.
Properties The density of water is taken to be 62.4 lbm/ft^3 (Table A-3E).
Analysis Applying the hydrostatic equation,
= = 1.44 psia
2
2
2 2
3 2
1.44lbf/in
144 in
1 ft
- 174 lbmft/s
1 lbf ( 62. 4 lbm/ft )(32.174ft/s )(40/12ft)
P ρ gh
1-49 The pressure given in mm Hg unit is to be converted to kPa.
Analysis Using the mm Hg to kPa units conversion factor,
⎟⎟= 133.3^ kPa ⎠
1 mmHg
0.1333kPa P ( 1000 mmHg)
PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
1-52 The gage pressure in a liquid at a certain depth is given. The gage pressure in the same liquid at a different depth is to be determined.
Assumptions The variation of the density of the liquid with depth is negligible.
Analysis The gage pressure at two different depths of a liquid can be expressed as
P 1 = ρ gh 1 and P 2 = ρ gh 2
h 1
(^1) h 2
Taking their ratio,
1
2 1
2 1
2 h
h gh
gh P
P
Solving for P 2 and substituting gives
= = ( 28 kPa)= 84 kPa 3 m
9 m 1 1
2 (^2) h P
h P
Discussion Note that the gage pressure in a given fluid is proportional to depth.
1-53 The absolute pressure in water at a specified depth is given. The local atmospheric pressure and the absolute pressure at the same depth in a different liquid are to be determined.
Assumptions The liquid and water are incompressible.
Properties The specific gravity of the fluid is given to be SG = 0.85. We take the density of water to be 1000 kg/m^3. Then density of the liquid is obtained by multiplying its specific gravity by the density of water,
SG (0.85)(100 0 kg/m^3 ) 850 kg/m^3 ρ= × ρ H 2 O = =
Analysis ( a ) Knowing the absolute pressure, the atmospheric pressure can be determined from P atm
h
= 96.0 kPa P
2
3 2
atm
1000 N/m
1 kPa (145kPa) (1000kg/m )(9.81m/s )(5m)
P P ρ gh
( b ) The absolute pressure at a depth of 5 m in the other liquid is
= 137.7 kPa
2
3 2
atm
1000 N/m
1 kPa (96.0kPa) (850kg/m )(9.81m/s )(5m)
P P ρ gh
Discussion Note that at a given depth, the pressure in the lighter fluid is lower, as expected.
PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
1-54E It is to be shown that 1 kgf/cm^2 = 14.223 psi.
Analysis Noting that 1 kgf = 9.80665 N, 1 N = 0.22481 lbf, and 1 in = 2.54 cm, we have
- 20463 lbf 1 N
0.22481lbf 1 kgf 9.80665N (9.80665N) ⎟⎟= ⎠
and
⎟⎟ = = 14.223^ psi ⎠
= =^2
2 (^2 2214). 223 lbf/in 1 in
2.54cm 1 kgf/cm 2. 20463 lbf/cm ( 2. 20463 lbf/cm )
1-55E The pressure in chamber 3 of the two-piston cylinder shown in the figure is to be determined.
Analysis The area upon which pressure 1 acts is
2
2 2 1 1 4 7.^069 in
( 3 in) 4
D
A
F 1
F 2
F 3
and the area upon which pressure 2 acts is
2
2 2 2 2 4 3.^142 in
( 2 in) 4
=π =π =
D
A
The area upon which pressure 3 acts is given by
2 A 3 = A 1 − A 2 = 7. 069 − 3. 142 = 3. 927 in
The force produced by pressure 1 on the piston is then
( 7. 069 in ) 1060 lbf 1 psia
1 lbf/in ( 150 psia)^2
2 (^1 11) ⎟⎟ = ⎠
F = PA =
while that produced by pressure 2 is
F 1 = P 2 A 2 =( 200 psia)( 3. 142 in^2 )= 628 lbf
According to the vertical force balance on the piston free body diagram
F 3 = F 1 − F 2 = 1060 − 628 = 432 lbf
Pressure 3 is then
= = 2 = 110 psia 3
3 3
- 927 in
432 lbf A
F
P
PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
1-58 The mass of a woman is given. The minimum imprint area per shoe needed to enable her to walk on the snow without sinking is to be determined.
Assumptions 1 The weight of the person is distributed uniformly on the imprint area of the shoes. 2 One foot carries the entire weight of a person during walking, and the shoe is sized for walking conditions (rather than standing). 3 The weight of the shoes is negligible.
Analysis The mass of the woman is given to be 70 kg. For a pressure of 0.5 kPa on the snow, the imprint area of one shoe must be
⎟= 1.37m^2 ⎠
2 2
2
1000 N/m
1 kPa 1 kgm/s
1 N
0.5kPa
(70kg)(9.81m/s )
P
mg P
W
A
Discussion This is a very large area for a shoe, and such shoes would be impractical to use. Therefore, some sinking of the snow should be allowed to have shoes of reasonable size.
1-59 The vacuum pressure reading of a tank is given. The absolute pressure in the tank is to be determined.
Properties The density of mercury is given to be ρ = 13,590 kg/m^3.
Analysis The atmospheric (or barometric) pressure can be expressed as
P abs
P atm = 750 mmHg
15 kPa
- 0 kPa
1000 N/m
1 kPa 1 kg m/s
1 N
(13,590 kg/m^3 )(9.807m/s^2 )(0.750m) 2 2
atm
P = ρ gh
Then the absolute pressure in the tank becomes
P abs = P atm− P vac=100.0− 15 = 85.0 kPa
1-60E A pressure gage connected to a tank reads 50 psi. The absolute pressure in the tank is to be determined.
Properties The density of mercury is given to be ρ = 848.4 lbm/ft^3.
Analysis The atmospheric (or barometric) pressure can be expressed as
P abs 50 psi
14.29 psia
144 in
1 ft 32.2lbmft/s
1 lbf (848.4 lbm/ft )(32.2ft/s )(29.1/12ft) 2
2 2
3 2
atm
P = ρ gh
Then the absolute pressure in the tank is
P abs = P gage+ P atm= 50 +14.29= 64.3 psia
PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
1-61 A pressure gage connected to a tank reads 500 kPa. The absolute pressure in the tank is to be determined.
P abs
P atm = 94 kPa
500 kPa
Analysis The absolute pressure in the tank is determined from
P abs = P gage+ P atm= 500 + 94 = 594 kPa
1-62 A mountain hiker records the barometric reading before and after a hiking trip. The vertical distance climbed is to be determined.
h =?
780 mbar
930 mbar
Assumptions The variation of air density and the gravitational acceleration with altitude is negligible.
Properties The density of air is given to be ρ = 1.20 kg/m^3.
Analysis Taking an air column between the top and the bottom of the mountain and writing a force balance per unit base area, we obtain
(0.930 0.780) bar 100,000N/m
1 bar 1 kg m/s
1 N
(1.20kg/m )(9.81m/s )()
2 2
3 2
air bottom top
air bottom top
h
gh P P
W A P P
It yields
h = 1274 m
which is also the distance climbed.
PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
1-64 EES Problem 1-63 is reconsidered. The entire EES solution is to be printed out, including the numerical results with proper units.
Analysis The problem is solved using EES, and the solution is given below.
P_bottom=755 [mmHg] P_top=730 [mmHg] g=9.807 [m/s^2] "local acceleration of gravity at sea level" rho=1.18 [kg/m^3] DELTAP_abs=(P_bottom-P_top)CONVERT('mmHg','kPa')"[kPa]" "Delta P reading from the barometers, converted from mmHg to kPa." DELTAP_h =rhogh/1000 "[kPa]" "Equ. 1-16. Delta P due to the air fluid column height, h, between the top and bottom of the building." "Instead of dividing by 1000 Pa/kPa we could have multiplied rhog*h by the EES function, CONVERT('Pa','kPa')" DELTAP_abs=DELTAP_h
SOLUTION Variables in Main DELTAP_abs=3.333 [kPa] DELTAP_h=3.333 [kPa] g=9.807 [m/s^2] h=288 [m] P_bottom=755 [mmHg] P_top=730 [mmHg] rho=1.18 [kg/m^3]
1-65 A diver is moving at a specified depth from the water surface. The pressure exerted on the surface of the diver by water is to be determined.
Assumptions The variation of the density of water with depth is negligible.
Properties The specific gravity of seawater is given to be SG = 1.03. We take the density of water to be 1000 kg/m^3.
Analysis The density of the seawater is obtained by multiplying its specific gravity by the density of water which is taken to be 1000 kg/m^3 :
P atm
Sea h
P
SG (1.03)(100 0 kg/m^3 ) 1030 kg/m^3 ρ= × ρ H 2 O = =
The pressure exerted on a diver at 30 m below the free surface of the sea is the absolute pressure at that location:
= 404.0 kPa
2
3 2
atm
1000 N/m
1 kPa (101kPa) (1030kg/m )(9.807m/s )(30m)
P P ρ gh
PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
1-66 A gas contained in a vertical piston-cylinder device is pressurized by a spring and by the weight of the piston. The pressure of the gas is to be determined.
Analysis Drawing the free body diagram of the piston and balancing the vertical forces yield
F spring
P atm
P
W = mg
PA = P atm A + W + F spring
Thus,
= 123.4 kPa
×
− 4 2 2
2
spring atm
1000 N/m
1 kPa 35 10 m
(4kg)(9.81m/s) 60 N (95kPa)
A
mg F P P
1-67 EES Problem 1-66 is reconsidered. The effect of the spring force in the range of 0 to 500 N on the pressure inside the cylinder is to be investigated. The pressure against the spring force is to be plotted, and results are to be discussed.
Analysis The problem is solved using EES, and the solution is given below.
g=9.807 [m/s^2] P_atm= 95 [kPa] m_piston=4 [kg] {F_spring=60 [N]} A=35CONVERT('cm^2','m^2')"[m^2]" W_piston=m_pistong"[N]" F_atm=P_atmACONVERT('kPa','N/m^2')"[N]" "From the free body diagram of the piston, the balancing vertical forces yield:" F_gas= F_atm+F_spring+W_piston"[N]" P_gas=F_gas/A*CONVERT('N/m^2','kPa')"[kPa]"
0 100 200 300 400 500
100
120
140
160
180
200
220
240
260
F spring [N]
P
gas
[kPa]
Fspring [N] Pgas [kPa] 0 106. 55.56 122. 111.1 138 166.7 153. 222.2 169. 277.8 185. 333.3 201. 388.9 217. 444.4 233. 500 249.
PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
